Find $\dfrac{d}{dx}\left[\dfrac{(\ln(x))^3}{x^3}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\ln(x)}{x^3}$ (Choice B) B $\dfrac{3[\ln(x)]^2\Bigl(x-\ln(x)\Bigr)}{x^4}$ (Choice C) C $\dfrac{3[\ln(x)]^2\Bigl(1-\ln(x)\Bigr)}{x^4}$ (Choice D) D $\dfrac{\ln(x)}{x^2}$
Explanation: $\dfrac{(\ln(x))^3}{x^3}$ is a quotient of a composite function and another function. Let... $u(x)=x^3$ $v(x)=\ln(x)$ $w(x)=x^3$... then $\dfrac{(\ln(x))^3}{x^3}=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $\dfrac{d}{dx}\left[ \dfrac{(\ln(x))^3}{x^3}\right]$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=3x^2$ $v'(x)=\dfrac1x$ $w'(x)=3x^2$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{3[\ln(x)]^2}\left(\dfrac1x\right)\left(x^3\right)-{[\ln(x)]^3}\cdot 3x^2}{(x^3)^2} \\\\ &=\dfrac{3x^2[\ln(x)]^2-3x^2[\ln(x)]^3}{x^6} \\\\ &=\dfrac{3x^2[\ln(x)]^2\left(1-\ln(x)\right)}{x^6} \\\\ &=\dfrac{3[\ln(x)]^2\left(1-\ln(x)\right)}{x^4}&&\gray{\text{Cancel common factors}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[ \dfrac{(\ln(x))^3}{x^3}\right]=\dfrac{3[\ln(x)]^2\left(1-\ln(x)\right)}{x^4}$.